Chapter 7 - Principles Of Integral Evaluation - 7.3 Integrating Trigonometric Functions - Exercises Set 7.3 - Page 507: 29

$$\frac{{{{\tan }^3}x}}{3} + C$$

$$\eqalign{ & \int {{{\tan }^2}x{{\sec }^2}x} dx \cr & {\text{substitute }}u = \tan x,{\text{ }}du = {\sec ^2}xdx \cr & = \int {{u^2}du} \cr & {\text{find the antiderivative by the power rule}} \cr & = \frac{{{u^3}}}{3} + C \cr & {\text{write in terms of }}x,{\text{ replace }}u = \tan x \cr & = \frac{{{{\tan }^3}x}}{3} + C \cr} $$